Answer to Question #307888 in Physics for cols

Question

A stationary proton is accelerated through a potential difference of 83.5 kV. What speed did the proton acquire?

Answer 1

The work done by the field on proton is by definition:

$$W = eV$$

where $e = 1.60\times10^{-19}C$ is the charge of proton, and $V = 86.5\times 10^3V$ is the potential difference. According to the work-energy theorem, this work is equal to the kinetic energy of the proton:

$$W = K= \dfrac{mv^2}{2}$$

where $m = 1.67\times 10^{-27}kg$ is the mass of proton, and $v$ is its speed. Thus, obtain:

$$\dfrac{mv^2}{2} = eV\\ v = \sqrt{\dfrac{2eV}{m}} = \sqrt{\dfrac{2\cdot 1.6\times 10^{-19}\cdot 86.5\times 10^3}{1.67\times 10^{-27}}} \approx 4.07\times 10^6m/s$$

Answer. $4.07\times 10^6m/s$